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1. Vector question
http://store.aqa.org.uk/qual/gceasa/...W-QP-JUN08.PDF

Q7. Last part.
Is it wrong to assume that AB is perpendicular to AC? I tried to use this in my solution (scalar product of ab and ac is 0), but got the wrong answer.
I got (2+x,5,-1-3x)"AC".(2,5,-1)"AB"=0 and therefore x=-6.
2. Re: Vector question
You are given that AB=AC so triangle ABC is isosceles. Angle BAC is only going to be 90 degrees if the angle between AN and L is 45 degrees. It isn't 45 degrees is it?
3. Re: Vector question
(Original post by Dog4444)
http://store.aqa.org.uk/qual/gceasa/...W-QP-JUN08.PDF

Q7. Last part.
Is it wrong to assume that AB is perpendicular to AC? I tried to use this in my solution (scalar product of ab and ac is 0), but got the wrong answer.
I got (2+x,5,-1-3x)"AC".(2,5,-1)"AB"=0 and therefore x=-6.
Distance of AB: V(2^2+5^2+(-1)^2)=V30
For AC: V(2+x)^2+25+(-1-3x)^2
Take equal and solve for x
10x^2+10x+30=30 -> 10x(x+1)=0
4. Re: Vector question
(Original post by BabyMaths)
You are given that AB=AC so triangle ABC is isosceles. Angle BAC is only going to be 90 degrees if the angle between AN and L is 45 degrees. It isn't 45 degrees is it?
What is AN and L?
Angles ABC and BCA are 45 degrees, because tan (angle)=1, isn't it?
5. Re: Vector question
(Original post by Dog4444)
What is AN and L?
Angles ABC and BCA are 45 degrees, because tan (angle)=1, isn't it?
Sorry. Typo. I meant AB and L.

What did you get for part b?
6. Re: Vector question
(Original post by BabyMaths)
Sorry. Typo. I meant AB and L.

What did you get for part b?
What is L? Eh? I got the right answer for the first two parts(73 for b).
7. Re: Vector question
(Original post by Dog4444)
What is L? Eh? I got the right answer for the first two parts(73 for b).
The line l.

You have the correct angle (73.22...) so you know that the angle BAC is not 90 degrees.
8. Re: Vector question
(Original post by BabyMaths)
The line l.

You have the correct angle (73.22...) so you know that the angle BAC is not 90 degrees.
Oh, ok. What it would look like than?
I drew it like that:

9. Re: Vector question
(Original post by Dog4444)
Oh, ok. What it would look like than?
I drew it like that:

It's OK but it doesn't really matter. The point is it's not a right angle. Maybe take a look at Ztibor's post now? His 'V' stands for square root.
10. Re: Vector question
Ok, there is another one:
http://store.aqa.org.uk/qual/gceasa/...W-QP-JAN09.PDF
Q8 last part.
Is it wrong to assume OD.CD=0? OD=(6+x,2,-4-x) and CD=(x,0,x)
But scalar producting these gave me x=0 what is wrong again.
11. Re: Vector question
(Original post by Dog4444)
Is it wrong to assume OD.CD=0?
That would not be an assumption. It says in the question angle ODC = 90 degrees.

Why are you using x,0,x for the direction of CD?

There are two problems with this. The variable is not helping and the correct direction is .

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